## Basis for a Topology

July 2020

**Problem 4:**

(a) If \(\{\mathscr{T}_{\alpha}\}\) is a family of topologies on \(X\), show that \(\bigcap \mathscr{T}_{\alpha}\) is a topology on \(X\). Is the same true for \(\bigcup \mathscr{T}_{\alpha}\)?

(b) Let \(\{\mathscr{T}_{\alpha}\}\) be a family of topologies on \(X\). Show that there is a unique smallest topology on \(X\) containing all collections \(\mathscr{T}_{\alpha}\), and a unique largest topology contained in all \(\mathscr{T}_{\alpha}\).

(a) All we must do is show that the intersection of the topologies has the properties of a topology:

\(\varnothing\) and \(X\) are clearly in \(\bigcap \mathscr{T}_{\alpha}\) since they are both in every \(\mathscr{T}_{\alpha}\) by definition.

Let \(x, y \in \bigcap \mathscr{T}_{\alpha}\). Since both \(x\) and \(y\) are contained in every \(\mathscr{T}_{\alpha}\), their union is also contained in every \(\mathscr{T}_{\alpha}\) by definition.

Since both \(x\) and \(y\) are contained in every \(\mathscr{T}_{\alpha}\), their intersection is also contained in every \(\mathscr{T}_{\alpha}\) by definition.

The same is not true for \(\bigcup \mathscr{T}_{\alpha}\) unless there is one \(\mathscr{T}_{\alpha}\) that contains all the other topologies. For example, suppose you have \(x \in \mathscr{T}_{\alpha}\) and \(y \in \mathscr{T}_{\beta}\). Unless \(y \in \mathscr{T}_{\alpha}\) or \(x \in \mathscr{T}_{\beta}\), \(x \cup y\) isn't necessarily in \(\mathscr{T}_{\alpha} \cup \mathscr{T}_{\beta}\), and so the second condition for a topology is not satisfied.

(b) The unique largest topology contained in all
\(\mathscr{T}_{\alpha}\) would be
\(\bigcap \mathscr{T}_{\alpha}\), since it is by
definition the largest set which is contained in every
\(\mathscr{T}_{\alpha}\), and it is itself a topology,
as proved above.

The unique smallest topology containing all
\(\mathscr{T}_{\alpha}\), which we shall call \(T\), is the

topology generated
by the subbasis \(\bigcup \mathscr{T}_{\alpha}\).

By definition, \(T\) would contain every \(x \in
\mathscr{T}_{\alpha} \forall \alpha\), every possible finite
intersection of \(\bigcup \mathscr{T}_{\alpha}\), and
every possible union of the finite intersections.

Now we are left to prove that this is in fact the smallest
topology containing every \(\mathscr{T}_{\alpha}\).

Suppose \( T'\) is a topology that is smaller than \(T\) and contains all of \(\{\mathscr{T}_{\alpha}\}\). Then there must be elements of \(T\) which are not in \(T'\). If \(T'\) does not contain any \(x \in \mathscr{T}_{\alpha}\) for any \(\alpha\), then it does not contain all of \(\{\mathscr{T}_{\alpha}\}\). If it does not contain some finite intersection of the elements of \(\bigcup \mathscr{T}_{\alpha}\), then it is not a topology by definition. If it does not contain some union of the finite intersections of \(\bigcup \mathscr{T}_{\alpha}\), then it is not a topology by definition. Thus, \(T\) must be the smallest topology which contains all of \(\{\mathscr{T}_{\alpha}\}\).

\(\tag*{$\blacksquare$}\)